What is the solution for y^{''}-5y^'=x^2-2x ?

The solution for y^{''}-5y^'=x^2-2x is y=c_{1}e^{5x}+c_{2}-(x^3)/(15)+(4x^2)/(25)+(8x)/(125)

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y^{''}-5y^'=x^2-2x

y^{^{''}} - 5 y^{^{'}} = x^{2} - 2 x

y = c_{1} e^{5 x} + c_{2} - \frac{x ^{3}}{15} + \frac{4 x ^{2}}{25} + \frac{8 x}{125}

Solution steps

y^{''} (x) - 5 y^{'} (x) = x^{2} - 2 x

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} - \frac{x ^{3}}{15} + \frac{4 x ^{2}}{25} + \frac{8 x}{125}

y = c_{1} e^{5 x} + c_{2} - \frac{x ^{3}}{15} + \frac{4 x ^{2}}{25} + \frac{8 x}{125}

y^{^{''}} - 2 y^{^{'}} + y = cos (2 x)

y^{^{''}} = y + e^{2 x} e^{e^{x}}

y^{^{''}} + 4 y^{^{'}} = t^{2} + 2 e^{t}

y^{^{'}} + y^{^{''}} + y = x e^{x}

y^{^{''}} + 2 y^{^{'}} - 15 y = (2 x + 1) e^{3 x} + 10

What is the solution for y^{''}-5y^'=x^2-2x ?
The solution for y^{''}-5y^'=x^2-2x is y=c_{1}e^{5x}+c_{2}-(x^3)/(15)+(4x^2)/(25)+(8x)/(125)